Electrochemistry+Capacitors+Current Electricity

Fundamental capacitor equation:

$\large{Q = C V}$

Rate of change of charge:

$\large{\dot{Q} = C \dot{V} + V \dot{C} = V \dot{C}}$

Capacitance formula:

$\large{C = \frac{K \epsilon_0 A}{d} \\ \dot{C} = - \frac{K \epsilon_0 A}{d^2} \dot{d}}$

Plugging into charge rate of change:

$\large{\dot{Q} = - \frac{V K \epsilon_0 A}{d^2} \dot{d} \\ \dot{d} = 2 a t + b }$

Total charge transfered (only magnitude matters):

$\large{\Delta Q = \int_0^2 \frac{V K \epsilon_0 A}{d^2} \dot{d} \, dt \\ = V K \epsilon_0 A \int_0^2 \frac{\dot{d}}{d^2} \, dt }$

We get one mole of silver for every mole of electrons. The number of moles of silver at the cathode is therefore:

$\large{N = \frac{\Delta Q}{F}}$

There are approximately $108$ grams for each mole of silver. The total mass, in grams, is therefore:

$\large{M_{tot} = 108 \frac{\Delta Q}{F} \\ = \frac{108 \, V K \epsilon_0 A}{F} \int_0^2 \frac{\dot{d}}{d^2} \, dt }$

Evaluating the integral numerically and plugging in numbers yields a total mass of $18 \epsilon_0$ grams.